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\title{\heiti\zihao{2} 习题3.4}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求下列无穷小或无穷大的阶:}
\subsection{$x-3 x^{2}+x^{6}(x \rightarrow 0)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{x-3 x^{2}+x^{6}}{x}=\lim _{x \rightarrow 0}\frac{1-3 x+x^{5}}{1}=1$$

所以其为$1$阶无穷小

\subsection{$\sqrt[5]{3 x^{2}-4 x^{3}}(x \rightarrow 0)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{\sqrt[5]{3 x^{2}-4 x^{3}}}{x^{\frac{2}{5}}}=\lim _{x \rightarrow 0} \frac{\sqrt[5]{3-4 x}}{1}=\lim _{x \rightarrow 0} \frac{\sqrt[5]{3}}{1}$$

所以其为$\frac{2}{5}$阶无穷小

\subsection{$x^{3}-3 x+2(x \rightarrow 1)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 1} \frac{x^{3}-3 x+2}{(x-1)^{2}}=\lim _{x \rightarrow 1}\frac{(x-1)^{2}(x+2)}{(x-1)^{2}}=\lim _{x \rightarrow 1}x+2=3$$

所以其为$2$阶无穷小

\subsection{$\sqrt{1+\tan x}-\sqrt{1-\sin x}(x \rightarrow 0)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim _{x \rightarrow 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}
		 & =\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{\frac{1}{2}}-(1+\sin x)^{\frac{1}{2}}\right]\left[(1+\tan x)^{\frac{1}{2}}+(1+\sin x)^{\frac{1}{2}}\right]}{x^{3}\left[(1+\tan x)^{\frac{1}{2}}+(1+\sin x)^{\frac{1}{2}}\right]}
		\\
		 & =\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{2 x^{3}}                                                                                                                                                                            \\
		 & =\lim _{x \rightarrow 0} \frac{\tan x(1-\cos x)}{2 x^{3}}                                                                                                                                                                         \\
		 & =\lim _{x \rightarrow 0} \frac{1-\cos x}{2 x^{2}}                                                                                                                                                                                 \\
		 & =\frac{1}{4} \lim _{x \rightarrow 0} \frac{1-\cos x}{\frac{1}{2} x^{2}}                                                                                                                                                           \\
		 & =\frac{1}{4}
	\end{aligned}
$$

所以其为$3$阶无穷小

\subsection{$\sqrt{1+\sqrt{1+\sqrt{x}}}-\sqrt{2}\left(x \rightarrow 0^{+}\right)$}
\textbf{解}\quad
$$\sqrt{1+\sqrt{1+\sqrt{x}}}-\sqrt{2}=\frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{1+\sqrt{x}}}+\sqrt{2}}=\frac{\sqrt{x}}{(\sqrt{1+\sqrt{1+\sqrt{x}}}+\sqrt{2})(\sqrt{1+\sqrt{x}}+1)}$$
$$\lim _{x \rightarrow 0} \frac{\sqrt{x}}{\frac{(\sqrt{1+\sqrt{1+\sqrt{x}}+\sqrt{2})(\sqrt{1+\sqrt{x}+1)}}}{x^{\frac{1}{2}}}}=\frac{1}{4 \sqrt{2}}$$

所以其为$\frac{1}{2}$阶无穷小

\subsection{$\sqrt{x+\sqrt{x+\sqrt{x}}}\left(x \rightarrow 0^{+}\right)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}}=\frac{\sqrt{1+\sqrt{o(1)}}}{1}=1$$

所以其为$\frac{1}{2}$阶无穷小

\subsection{$x^{x}-1(x \rightarrow 1)$}
\textbf{解}\quad
$$\lim _{x \rightarrow 1} \frac{e^{x \ln x}-1}{x-1}=\lim _{x \rightarrow 1}\frac{x \ln x}{x-1}=1$$

所以其为$1$阶无穷小

\subsection{$x-x^{2}+3 x^{8}(x \rightarrow \infty)$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty} \frac{3 x^{8}-x^{2}+x}{x^{8}}=\frac{3-\frac{1}{x^{6}}+\frac{1}{x^{7}}}{1}=3$$

其为$8$阶无穷大

\subsection{$\frac{x-2}{x^{4}+1}(x \rightarrow \infty)$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty} x^{3} \cdot \frac{x-2}{x^{4}+1}=\lim _{x \rightarrow \infty} \frac{1-\frac{2}{x}}{1+\frac{1}{x^{4}}}=1$$

所以其为$3$阶无穷小.

\subsection{$(1+x)\left(1+x^{2}\right) \cdots\left(1+x^{n}\right)(x \rightarrow \infty)$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty} \frac{\prod_{i=1}^{n}\left(1+x^{i}\right)}{x^{\frac{n(n+1)}{2}}}=\frac{1+O\left(\frac{1}{x}\right)}{1}=1$$

其为$\frac{n(n+1)}{2}$阶无穷大.

\section{当 $x \rightarrow 0$ 时，下列式子成立吗?}
\subsection{$o\left(x^{2}\right)=o(x)$}
\textbf{解}\quad
不正确,其意义是$o(x^2) \subset o(x)$.

\subsection{$O\left(x^{2}\right)=o(x)$}
\textbf{解}\quad
不正确,其意义是$O(x^2) \subset o(x)$

\subsection{$x o\left(x^{2}\right)=o\left(x^{3}\right)$}
\textbf{解}\quad
正确,左右两边都是比三阶无穷小量更小的小量.

\subsection{$\frac{o\left(x^{2}\right)}{x}=o(x)$}
\textbf{解}\quad
正确,左右两边都代表比一阶无穷小量更小的小量.

\subsection{$\frac{o\left(x^{2}\right)}{o(x)}=o(x)$}
\textbf{解}\quad
不正确,因为比某阶小量更小的小量可以是任意小的阶的小量,因为其不确定,所以不能直接四则运算

\textbf{\textcolor{red}{注}}\quad
$O(x^a)$意义是$x$是$a$阶无穷小或更低阶的小量.$o(x^a)$的意义是$x$是比$a$阶无穷小量更小的无穷小量.

\section{分别求满足下列条件的常数 $a, b$ :}
\subsection{$\lim _{x \rightarrow+\infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow+\infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=\lim _{x \rightarrow+\infty}\frac{(1-a) x^{2}-(a+b) x+(1-b)}{x+1}=0
$$

显然$a=1,a+b=0$,所以$a=1,b=-1$.

\subsection{$\lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}-x+1}-a x-b\right)=0$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}-x+1}-a x-b\right)=\lim _{x \rightarrow-\infty}\frac{\left(1-a^{2}\right) x^{2}-(1+2 a b) x+\left(1-b^{2}\right)}{\sqrt{x^{2}-x+1}+a x+b} =0
$$

所以$1-a^2=0,1+2ab=0$;而显然$a<0$否则极限不可能为$0$,所以a$=-1,b=\frac{1}{2}$.

\section{用等价无穷小替换求下列极限:}
\subsection{$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \tan x}-1}{1-\cos x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\sqrt{1+x \tan x}-1}{1-\cos x}=\lim _{x \rightarrow 0}\frac{\frac{1}{2} x \tan x}{\frac{1}{2} x^{2}}=1
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\mathrm{e}^{\sin x}-1}{\arctan (2 x)}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\arctan (2 x)}=\lim _{x \rightarrow 0} \frac{\sin x}{2 x}=\frac{1}{2}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{(1+x)^{x}-1}{x \arcsin x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow \infty} \frac{(1+x)^{x}-1}{x \arcsin x}=\lim _{x \rightarrow 0} \frac{x \cdot x}{x^{2}}=1
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\ln \cos a x}{\ln \cos b x}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\ln \cos a x}{\ln \cos b x}=\lim _{x \rightarrow 0} \frac{\ln (1+(\cos a x-1))}{\ln (1+(\cos b x-1))}=\lim _{x \rightarrow 0} \frac{\cos a x-1}{\cos b x-1}=\frac{-\frac{1}{2}(a x)^{2}}{-\frac{1}{2}(b x)^{2}}=\frac{a^{2}}{b^2}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-1}{\mathrm{e}^{x^{2}}-1}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-1}{e^{x^{2}}-1}=\lim _{x \rightarrow 0}\frac{x \sin x}{2x^{2}}=\frac{1}{2}
$$

\subsection{$\lim _{x \rightarrow 0} \frac{\left(1+x^{2}-2 x^{3}\right)^{\frac{1}{n}}-1}{\cos x-1}, n \in \mathbb{N}^{*}$}
\textbf{解}\quad
$$
	\lim _{x \rightarrow 0} \frac{\left(1+x^{2}-2 x^{3}\right)^{\frac{1}{n}}-1}{\cos x-1}=\lim _{x \rightarrow 0} \frac{\frac{1}{n}\left(x^{2}-2 x^{3}\right)}{-\frac{1}{2} x^{2}}=\lim _{x \rightarrow 0}-\frac{2(1-2 x)}{n}=-\frac{2}{n}
$$

\section{计算下列极限 :}
\subsection{$\lim _{x \rightarrow \infty}\left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)^{x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow \infty}\left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)^{x}=\lim _{x \rightarrow \infty} e^{x \ln \left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)}$$
$$\lim _{x \rightarrow \infty} x \ln \left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)=\lim _{x \rightarrow \infty}x\left(\sin \frac{1}{x}+\cos \frac{1}{x}-1\right)=\lim _{t \rightarrow 0}\frac{\sin t+\cos t-1}{t}=\lim _{t \rightarrow 0}1-\frac{t}{2}=1$$
$$\therefore\lim _{x \rightarrow \infty}\left(\sin \frac{1}{x}+\cos \frac{1}{x}\right)^{x}=e$$

\subsection{$\lim _{x \rightarrow 0}\left(\frac{1+\sin x}{1+\tan x}\right)^{\frac{1}{x(1-\cos x)}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 0}\left(\frac{1+\sin x}{1+\tan x}\right)^{\frac{1}{x(1-\cos x)}}=\lim _{x \rightarrow 0}e^{\frac{1}{x(1-\cos x)} \cdot \ln \left(1+\frac{1+\sin x}{1+\tan x}-1\right)}=\lim _{x \rightarrow 0}e^{\frac{1}{x(1-\cos x)} \cdot\left(\frac{1+\sin x}{1+\tan x}-1\right)}$$
$$\lim _{x \rightarrow 0} \frac{1}{x(1-\cos x)}\left(\frac{1+\sin x}{1+\tan x}-1\right)
	=\lim _{x \rightarrow 0} \frac{\sin x-\tan x}{x(1-\cos x)(1+\tan x)}
	=\lim _{x \rightarrow 0} \frac{\tan x(\cos x-1)}{x(1-\cos x)}=-1$$
$$\therefore\lim _{x \rightarrow 0}\left(\frac{1+\sin x}{1+\tan x}\right)^{\frac{1}{x(1-\cos x)}}=e^{-1}$$

\subsection{$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\tan 2 x}$}
\textbf{解}\quad
$$\lim _{x \rightarrow \frac{\pi}{4}}(\tan x)^{\tan 2 x} =\lim _{x \rightarrow \frac{\pi}{4}}e^{\tan 2 x \ln \tan x} =\lim _{x \rightarrow \frac{\pi}{4}}e^{\tan 2 x \cdot(\tan x-1)}$$
$$
	\begin{aligned}
		\lim _{x \rightarrow \frac{\pi}{4}} \tan 2 x(\tan x-1) & =\lim _{x \rightarrow \frac{\pi}{4}} \tan 2 x \frac{(\sin x-\cos x)}{\cos x}                                               \\
		                                                       & =\lim _{x \rightarrow \frac{\pi}{4}} \tan 2 x \cdot \frac{\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)}{\cos x}              \\
		                                                       & =\lim _{x \rightarrow \frac{\pi}{4}}2 \sin \left(x-\frac{\pi}{4}\right) \cdot \frac{1}{\tan 2\left(\frac{\pi}{4}-x\right)} \\
		                                                       & =-1
	\end{aligned}
$$

\subsection{$\lim _{x \rightarrow 1}\left(3 \mathrm{e}^{\frac{x-1}{x+2}}-2\right)^{\frac{2 x^{2}}{x-1}}$}
\textbf{解}\quad
$$\lim _{x \rightarrow 1}\left(3 e^{\frac{x-1}{x+2}}-2\right)^{\frac{2 x^{2}}{x-1}}
	=\lim _{x \rightarrow 1} e^{\frac{2 x^{2}}{x-1}} \cdot \ln \left(1+3 e^{\frac{x-1}{x+2}}-3\right)\lim _{x \rightarrow 1} e^{\frac{2 x^{2}}{x-1}\left(3 e^{\frac{x-1}{x+2}}-3\right)}$$
$$\lim _{x \rightarrow 1} \frac{2 x^{2}}{x-1}\left(3 e^{\frac{x-1}{x+2}}-3\right)=\lim _{t \rightarrow 0} \frac{2(t+1)^{2}}{t}\left(3 e^{\frac{t}{t+3}}-3\right)=\lim _{t \rightarrow 0} \frac{(t+1)^{2}}{t} \cdot \frac{t}{t+3}\cdot6=2$$
$$\therefore\lim _{x \rightarrow 1}\left(3 e^{\frac{x-1}{x+2}}-2\right)^{\frac{2 x^{2}}{x-1}} =e^2$$




\end{document}
\subsection{}
\textbf{解}\quad

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\textbf{证}\quad